Answer:
Maximum at points (8,0),(-8,0).Minimum at points (0,8), (0,-8).
Step-by-step explanation:
There are multiple ways of using lagrange multipliers. Most of them are equivalent.
Consider the function
. We want the following
.
Then, we have
![\frac{\partial F}{\partial x} = 2x-2x\lambda= 2x(1-\lambda)=0](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cpartial%20F%7D%7B%5Cpartial%20x%7D%20%3D%202x-2x%5Clambda%3D%202x%281-%5Clambda%29%3D0)
![\frac{\partial F}{\partial y} = -2y-2y\lambda = -2y(1+\lambda)=0](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cpartial%20F%7D%7B%5Cpartial%20y%7D%20%3D%20-2y-2y%5Clambda%20%3D%20-2y%281%2B%5Clambda%29%3D0)
![\frac{\partial F}{\partial \lambda} = x^2+y^2-64=0](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cpartial%20F%7D%7B%5Cpartial%20%5Clambda%7D%20%3D%20x%5E2%2By%5E2-64%3D0)
From the first two equations, we can see that if
then necessarily y=0. IN that case, from the third equation (which is the restriction) gives us that
.
On the other hand, if
then necessarily x=0. Again, using the restriction this gives us that
.
if we evaluate the original function in this points, we have that
. Then, we have Maximum at points (8,0),(-8,0) and Minimum at points (0,8), (0,-8).